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Lesson 3 of 7

Finding Common Ground · Lesson 3 of 7

Finding the HCF from Prime Factors

“Match the shared prime pieces to find the greatest possible common factor.”

Learning Objectives

• Construct common factors from shared prime pieces. • Use the smallest count of each shared prime to obtain the HCF. • Find HCF for two or more numbers, including coprime pairs. • Explain why arbitrary composite factorizations can conceal shared factors.

The greatest shared subproduct

Every common factor must be buildable from the prime pieces of each number. For 45 = 3 × 3 × 5 and 75 = 3 × 5 × 5, both have one 3 and one 5 in common. Their shared products are 1, 3, 5 and 15, so the highest common factor is 15. We can find it without listing every factor.

72 = 2 × 2 × 2 × 3 × 330 = 2 × 3 × 5common: 2 × 3 = 6all needed: 2³ × 3² × 5HCF uses the smaller countLCM uses the larger count
Two different prime counts— The HCF uses the smaller count; the LCM comparison will come later.

For each prime, compare how often it appears in all the numbers. Include it only as many times as it appears in the factorisation with the fewest copies. Ignore primes missing from any one of them. This produces the greatest factor that divides all the numbers.

Example of the HCF ruleLaTeX
Use one 2 and one 3 because those are the smaller occurrence counts in the two prime factorisations.
Example — A repeated shared prime

Problem
Find the HCF of 112 and 84.

  1. 1.112 = 2 × 2 × 2 × 2 × 7; 84 = 2 × 2 × 3 × 7.
  2. 2.The smaller count of 2 is two, and both numbers have one 7.
  3. 3.HCF = 2 × 2 × 7 = 28. Check: 112 ÷ 28 = 4 and 84 ÷ 28 = 3.
Example — No shared prime

Problem
Find the HCF of 96 and 275.

  1. 1.96 = 2⁵ × 3 and 275 = 5² × 11.
  2. 2.There is no common prime factor, so the only shared factor is 1.
  3. 3.HCF = 1. Such a pair is called coprime even though neither number has to be prime.
Example — A larger pair

Problem
Find the HCF of 225 and 750.

  1. 1.225 = 3² × 5²; 750 = 2 × 3 × 5³.
  2. 2.Choose one 3 and two 5s, the smaller counts.
  3. 3.HCF = 3 × 5² = 75.

Extend the same reasoning

With three numbers, the selected number of copies of a prime must fit in all three factorisations. For 42, 75 and 24, no prime occurs in every number, so their HCF is 1. A composite-factor split can hide a common prime: 72 = 6 × 12 and 144 = 8 × 18 look different, but prime factorisation reveals their many shared 2s and 3s.

Example — Uncover a hidden HCF

Problem
Find the HCF of 30 and 72.

  1. 1.30 = 2 × 3 × 5; 72 = 2³ × 3².
  2. 2.Take one 2 and one 3; 5 is not shared.
  3. 3.HCF = 2 × 3 = 6.
Composite factors are not the comparison units

Do not conclude that 72 and 144 have HCF 1 because 6, 12, 8 and 18 look different. Continue to primes, then compare occurrences.

Quiz

Quick check

What is the HCF of 45 = 3² × 5 and 75 = 3 × 5²?

Quick check

For the HCF, how many copies of 2 are taken from 2⁴ and 2²?

Quick check

What is the HCF of 96 and 275?

Quick check

The HCF of 30 and 72 is

Quick check

If a prime occurs in two of three numbers but not the third, how often can it occur in their HCF?

Practice Problems

Practice Problems
  1. Find the common factors and HCF of 50 and 60.
  2. Use prime factorizations to find the HCF of 370 and 592.
  3. Find the HCF of 42, 75 and 24; explain why.
  4. Find the HCF of 81 and 243 without listing every factor.
  5. Explain why 72 = 6 × 12 and 144 = 8 × 18 do not imply HCF 1.
  6. Find the HCF of 400 and 2500, showing the chosen prime counts.

Key Takeaways

Key Takeaways

• Every common factor comes from prime pieces present in every number. • For HCF, use the minimum number of copies of each shared prime. • Missing primes contribute no factor to the HCF. • HCF 1 means the numbers are coprime, not necessarily individually prime. • The same matching method extends to three or more numbers.

Previous · Lesson 2

Prime Factorisation and All the Factors

Next · Lesson 4

Common Multiples and the LCM